Question: Let $S$ be the quarter of the cylinder with height $1$ and radius $5$ such that $\pi < \theta < \dfrac{3\pi}{2}$, whose axis is parallel to the $z$ -axis and whose lower base is centered at the origin. What is the triple integral of the scalar field $f(x, y, z) = yz$ over $S$ in cylindrical coordinates? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^1 \int_0^5 \int_\pi^{\frac{3\pi}{2}} r^2z\sin(\theta) \, d\theta \, dr \, dz$ (Choice B) B $ \int_0^5 \int_0^1 \int_\pi^{\frac{3\pi}{2}} rz\cos(\theta) \, d\theta \, dr \, dz$ (Choice C) C $ \int_0^1 \int_0^5 \int_\pi^{\frac{3\pi}{2}} rz\cos(\theta) \, d\theta \, dr \, dz$ (Choice D) D $ \int_0^5 \int_0^1 \int_\pi^{\frac{3\pi}{2}} r^2z\sin(\theta) \, d\theta \, dr \, dz$
Solution: The only bound is $0 < \theta < 2\pi$. Here is the change of variables for cylindrical coordinates. $\begin{aligned} x &= r \cos(\theta) \\ \\ y &= r \sin(\theta) \\ \\ z &= z \end{aligned}$ We want to represent the cylinder $S$ with bounds in cylindrical coordinates. Here, the region $S$ ranges from a height of $z = 0$ to $z = 1$ and radius of $r = 0$ to $r = 5$. Theta goes from $\pi$ to $\dfrac{3\pi}{2}$ so that we only integrate over the specified quarter. $ \int_0^1 \int_0^5 \int_\pi^{\frac{3\pi}{2}} \cdots \, d\theta \, dr \, dz$ We can now put $f(x, y, z)$ in the integrand, but we need to substitute $x$, $y$, and $z$ for their definitions in cylindrical coordinates. $ \int_0^1 \int_0^5 \int_\pi^{\frac{3\pi}{2}} rz\sin(\theta) \cdots \, d\theta \, dr \, dz$ The final step is finding the Jacobian of cylindrical coordinates, which we'll need to multiply in to get the final integral. $J(r, \theta, z) = r$ [Derivation] The integral in cylindrical coordinates: $ \int_0^1 \int_0^5 \int_\pi^{\frac{3\pi}{2}} r^2z\sin(\theta) \, d\theta \, dr \, dz$